Let $a(x)=-7x^5+3x^3-6x-8$, and $b(x)=x^2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{-7x^5+3x^3-6x-8}{x^2}&=\dfrac{-7 {x^5}+3 {x^3}}{ {x^2}}+\dfrac{-6x-8}{x^2}\\\\ &={-7x^3+3x}+\dfrac{{-6x-8}}{x^2}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${-6x-8}$ is less than the degree of $x^2$, it follows that ${r(x)}={-6x-8}$, and ${q(x)}={-7x^3+3x}$. To conclude, $q(x)=-7x^3+3x$ $r(x)=-6x-8$ [Is there another way of doing this?]